# Chemical Reaction Engineering Questions and Answers – Rate Data Collection & Analysis – Method of Half Lives

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This set of Chemical Reaction Engineering Questions and Answers for Aptitude test focuses on “Rate Data Collection & Analysis – Method of Half Lives”.

1. State true or false.
For a second order reaction, t.0.5 is directly proportional to the initial concentration.
a) True
b) False

Explanation: The general expression relating t.0.5, rate constant and initial concentration is t.0.5 = $$\frac{C_{A0}^{1-n}(2^{n-1}-1)}{(n-1)k}.$$ For a second order reaction, t.0.5 = $$\frac{1}{kC_{A0}}.$$ Hence, t.0.5 is inversely proportional to the initial concentration.

2. In terms of conversion, half life is the time taken for ____
a) 50% conversion
b) 75% conversion
c) 20% conversion
d) 30% conversion

Explanation: Half Life is the time to reduce the reactant concentration to half of initial value. Hence, 50% of the reactant is converted to product.

3. For all orders of reaction except the first order, half life is related to initial concentration as ____
a) t0.5 α CAon
b) t0.5 α CAo1-n
c) t0.5 α CAo2n
d) t0.5 α CAo

Explanation: The generalised equation is t0.5 = $$\frac{C_{A0}^{1-n}(2^{n-1}- 1)}{(n-1)k}.$$ At half life time, for a reaction of order n, CA = 0.5CAo.

4. The plot representing the relationship between ln(t0.5) and ln(CAo) for second order reaction is ___
a) b) c) d) Explanation: The slope of the line is (1 – 2) = -1. Hence, the straight line is linearly decrementing.

5. The relationship between t.0.5 and CAo for a zero order reaction is ____
a) t0.5 = $$\frac{1}{kC_{A0}}$$
b) t0.5 = $$\frac{1}{C_{A0}}$$
c) t0.5 = $$\frac{C_{A0}}{k}$$
d) t0.5 = $$\frac{C_{A0}}{2k}$$

Explanation: For zero order reaction, $$\frac{-dC_A}{dt}$$ = k. t0.5 = $$\frac{C_{A0}- \frac{C_{A0}}{2}}{k}.$$
Hence, t0.5 = $$\frac{C_{A0}}{2k}.$$

6. For a zero order reaction, the slope of t0.5 and CAo is ____
a) 2
b) 3
c) 1
d) 0

Explanation: Slope is (1-n). At n=0, slope = 1.

7. What is the slope of the plot of ln (t0.5) and ln (CAo)? (Where n is the reaction order)
a) 1-n
b) 2-n
c) n
d) $$\frac{n}{2}$$

Explanation: The plot is a straight line of slope (1-n). The equation representing the relationship is ln (t0.5) = ln$$(\frac{(2^{n-1}-1)}{(n-1)k})$$ + (1 – n) ln (CAo).

8. If t0.5 = 100 minutes and initial concentration is 10, then the order of the reaction is ____
a) 4
b) -1
c) -2
d) 1

Explanation: t0.5 = $$C_{A0}^{1-n}.$$100 = 10(1 – n). Hence, n = -1.

9. The fractional conversion is expressed as____
a) $$\frac{C_A}{C_{A0}}$$
b) $$\frac{C_{A0}}{C_A}$$
c) $$\frac{1}{C_{A0}}$$
d) $$\frac{C_A}{kC_{A0}}$$

Explanation: Fractional conversion is the time required to reduce the concentration to any fraction. It is the ratio of final to initial concentration.

10. The time required to convert to a fraction of 0.7 for a first order reaction is ____
a) t0.7 = $$\frac{(0.7)}{C_{A0} k}$$
b) t0.7 = $$\frac{(-0.7)}{C_{A0} k}$$
c) t0.7 = $$\frac{(0.3)}{C_{A0} k}$$
d) t0.7 = $$\frac{(-0.3)}{C_{A0} k}$$

Explanation: The time for converting to any fraction, F is, tF = $$\frac{C_{A0}^{1-n}(F^{n-1}-1)}{(n-1)k}.$$ At F = 0.7, t0.7 = $$\frac{C_{A0}^{1-2}(0.7^1- 1)}{k}.$$ 